What is the $\text{p}K_\text{a}$ of water? A simple google search yields the value $15.74$, but this site and this paper say it's $14.0$.
According to my understanding, the correct answer should be $14.0$:
$$\text{p}K_{\text{a}}= -\log([\ce{H+}][\ce{OH-}])$$
For $25~\text{°C}$: $$[\ce{H+}][\ce{OH-}] = 10^{-14} = K_{\text{w}}$$
Thus follows $\text{p}K_{\text{a}}=14$.
Can you tell me which value is correct and why?
No comments:
Post a Comment